Reverse A Number

package jan18;

import java.util.Scanner;

public class reverseNumber {
    public static void main(String[] args) {

        // Input the number
        // Time Complexity: O(1) for reading input
        int n = new Scanner(System.in).nextInt();

        // Call method to reverse and print the number
        // Time Complexity: O(log10(n)) for reversing the digits
        method1(n);
    }

    // Method to reverse and print the number
    // Time Complexity (TC): O(log10(n)) in the worst case
    // O(log10(n)) can be simplified to O(log n) in Big O notation
    // - Best case (n = 0): O(1), no iteration
    // - Average case: O(log10(n)), proportional to the number of digits
    // - Worst case (largest n): O(log10(n)), processing all digits
    // Space Complexity (SC): O(1), constant space usage, O(1), constant space usage because
    // no additional data structures or variables are used apart from the input variable n.
    private static void method1(int n) {

        // Handle negative numbers
        // Time Complexity: O(1)
        if (n < 0) {
            n = -n;  // Convert negative number to positive
        }

        // Loop to reverse the digits
        // Time Complexity: O(log10(n)) because the number of digits is proportional to log10(n)
        while (n > 0) {
            // Extract the last digit and print it
            // Time Complexity: O(1) for modulo and printing
            System.out.print(n % 10);

            // Remove the last digit
            // Time Complexity: O(1) for division
            n = n / 10;
        }

    }
}

eg 1.

while (n > 1) {
    n = n / 2;
}

Here Each step halves n.
Total steps = log₂(n) → again O(log n)

eg 2.
while (n > 0) {
    System.out.print(n % 10);
    n = n / 10;
}
(Specifically, log₁₀(n) — but we simplify to O(log n) in Big O notation.)